The Intermediate Value Theorem is used frequently to establish that a solution to an equation exists (but oftern is no help in finding that solution, except by numeric approximation methods).

As an example: Show there is a solution to the equation cos(x) = x.

This is equivalent to showing that cos(x) - x = 0 is solvable.

Since f(x) = cos(x) - x is a continuous function and f(0) = 1 > 0 and f(pi/2) = -pi/2 < 0 there must exist some value c between 0 and pi/2 where f(c) = 0 (by the Intermediate Value Theorem), hence there is a solution to cos(x) = x.

Well, unless there is an integer root, that won't "solve" the equation! But that is a good way to start. In fact, it is exactly what jgens suggested. If x= 0 [itex]x^3+ x= 0+ 0= 0< 6[/itex]. If x= 2, [itex]x^3+ x= 8+ 2= 10> 6[/itex]. What does that tell you?

Well, unless there is an integer root, that won't "solve" the equation! But that is a good way to start. In fact, it is exactly what jgens suggested. If x= 0 [itex]x^3+ x= 0+ 0= 0< 6[/itex]. If x= 2, [itex]x^3+ x= 8+ 2= 10> 6[/itex]. What does that tell you?

It tells me that the solution is between 0 and 2, i.e [itex]x\in (0,2)[/itex].

Infact if you choose x=1, [itex]1^3+1=1+1=2<6[/itex], so [itex]x \in (1,2)[/itex]

If you go by checking for x=1.5 , [itex]1.5^3+1.5=4.875<6[/itex] you will come up with the solution. Now [itex]x \in (1.5 , 2)[/itex] and so on...